Last Updated on July 29, 2024 by Maged kamel

## Nominal shear strength for M10x7.50-use CM-15.

### Compute the Nominal shear strength for M10x7.50.

We have a given M10x7.50 steel section of A572 grade 65; it is required to compute the nominal shear strength. We will apply CM#15 and the related specification 2016 in this example.

First, we will use Table 2-4 to find the yield stress Fy value for A572 steel grade 65. This type of steel has a yield stress Fy=65 ksi and an ultimate stress Fu=80 ksi.

The next question concerns the table used to obtain the complete data for M10x7.5; in part 1, we will find that Tables 1-2 are used to get the information for sections.

The data for M10x7.50 Can be obtained from Table 1-2 from Part 1. The flange width bf is 2.69 inches, and the thickness is 0.173 inches.

The depth is d=9.99 inches, and the web thickness is 0.13 inches. The most critical parameter is h/tw, which we will find in part 2 of the same table.

### Find h/tw from Table 1-3, part 2.

The ratio for h over the thickness of the web is equal to 71. The distance h is between the web’s two filleted portions. The following slide image shows part 2 of Table 1-2.

### Find the limiting h/w based on Fy.

The limiting h/tw and the yield stress are related, as shown in the next slide image. Based on the Fy of the given M10x7.5 for grade 65, the limiting h/w equals 47.30.

The following slide image shows the relation of h/tw and the nominal shear value. There are three zones; the first zone is from zero to 47.30, marked as point 1 on the x-axis, which is obtained from the relation of 2.24*sqrt(E/Fy); in this zone, Cv1, unlive cv in Cm#14, equals 1, and the ฯv equals 1. The equation gives the nominal shear value for nominal shear Vn=Cv*0.60Fy*d*tw and can be written as 1*0.60*Fy*Aw.

The second Zone is from h/tw equals 47.30 to 53.692, marked as point 2 on the x-axis, and is obtained from the relation of 1.1*sqrt(kv *E/Fy); in this zone, Cv equals 1, and the ฯv equals 0.90. the Kv factor equals 5.34 in specification 2016. The nominal shear value is given by the equation Vn=Cv*0.60Fy*d*tw and can be written as 1*0.60Fy*Aw.

The third Zone is from h/tw, which equals 53.69 to the end; Cv1 is less than 1 in this zone, and the ฯv equals 0.90. The nominal shear value is given by the equation Vn=Cv1*0.60Fy*d*tw; Cv1 equals 53.69/h/tw. For our section, h/tw=71.0, and the value of cv1=53.69/71=0.756. Please refer to the next slide image.

### The nominal shear strength for M10x7.50.

The area of the web for M10x7.50 equals the product of d*tw or (9.99*0.13) or 1.2987 inch2; the cv1 value equals 0.7562, and the nominal shear equals 0.7562*(0.60*Fy)*(Aw)=0.7562*0.6*65*1.2987=38.30 kips.

### Excel plot for h/tw versus Vn.

I have used an Excel plot for the relation between h/tw, and the nominal shear Vn shows the zones for shear. Zone 1 starts h/tw value from 0 to 47.30, and the nominal value equals 50.65 Kips

. Zone 2 begins for h/tw from 47.30 to 53.69.and the nominal value equals 50.65 Kips. Zone 3 starts from h/tw bigger than 53.69; The Nominal shear Vn equals 50.41 kips for h/tw equals 71; the section has a nominal shear Vn value from 38.30 kips for h/tw=71.0.Thanks a lot.

As an external resource for the shear stress limit state.

The previous post is on nominal shear strength for S41x121.